What is the integral of (x^3)/(x^4-3) ?

The integral of (x^3)/(x^4-3) is 1/4 ln|x^4-3|+C

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integral of (x^3)/(x^4-3)

\int \frac{x ^{3}}{x ^{4} - 3} d x

\frac{1}{4} ln x^{4} - 3 + C

Solution steps

\int \frac{x ^{3}}{x ^{4} - 3} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = x^{4} - 3

= \frac{1}{4} ln x^{4} - 3

Add a constant to the solution

= \frac{1}{4} ln x^{4} - 3 + C

\int_{0}^{2} x^{2} - 1 d x

f (x) = - ln (x)

\int 5 x^{- 3} d x

\frac{d}{d y} (sin (y))

\int e^{- x} sin (e^{- x}) d x

What is the integral of (x^3)/(x^4-3) ?
The integral of (x^3)/(x^4-3) is 1/4 ln|x^4-3|+C